BZOJ4386 [POI2015]Wycieczki 矩阵+倍增

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=4386

题解

一眼就可以看出来是邻接矩阵快速幂。

可是这里的边权不为 \(1\)。不过可以发现，边权最多为 \(3\)。但是边的数量很多，不适合拆边，那就拆点吧。对于一条 \(x \to y\) 的边，就建立一个 \(x_0\to y_{w - 1}\) 的边，\(w\) 为边权。

然后就建立矩阵就可以了。因为我们需要统计第 \(i\) 步之前一共有多少路径，所以可以新建一个节点，每个点向这个点连一条有向边，这个点自己再来一个自环。

然后预处理 \(B_i\) 为走了 \(2^i\) 步的矩阵，直接倍增出来答案就可以了。

下面是代码，矩阵乘法的复杂度为 \(O(n^3)\)，一共倍增 \(O(\log k)\) 次，因此总的时间复杂度为 \(O(n^3\log k)\)。

#include<bits/stdc++.h>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b , 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b , 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I>
inline void read(I &x) {
    int f = 0, c;
    while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
    x = c & 15;
    while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
    f ? x = -x : 0;
}

const int N = 40 * 3 + 7;

int n, m;
ll k;

struct Matrix {
    ll a[N][N];
    inline Matrix() { memset(a, 0, sizeof(a)); }
    
    inline Matrix operator * (const Matrix &b) {
        Matrix c;
        for (int k = 0; k <= n; ++k)
            for (int i = 0; i <= n; ++i)
                for (int j = 0; j <= n; ++j)
                    c.a[i][j] += a[i][k] * b.a[k][j];
        return c;
    }
    
    inline void print() const {
        for (int i = 0; i <= n; ++i) {
            for (int j = 0; j <= n; ++j) dbg("%lld ", a[i][j]);
            dbg("\n");
        }
    }
} A, B[N];

inline bool isfull(const Matrix &a) {
    ll cnt = 0;
    for (int i = 1; i <= n / 3; ++i) {
        cnt += a.a[i][0] - 1;
        if (cnt >= k) return 1;
    }
    return 0;
}

inline void work() {
    n = n * 3, B[0] = A;
    int lim = 0;
    for (int i = 1; i <= 70; ++i) {
        B[i] = B[i - 1] * B[i - 1];
        ++lim;
        if (isfull(B[i])) break;
    }
    if (!isfull(B[lim--])) {
        puts("-1");
        return;
    }
    memset(A.a, 0, sizeof(A.a));
    for (int i = 0; i <= n; ++i) A.a[i][i] = 1;
    ll ans = 0;
    for (int i = lim; ~i; --i) {
        const Matrix &tmp = A * B[i];
        if (!isfull(tmp)) A = tmp, ans += 1ll << i;
    }
    printf("%lld\n", ans);
}

inline void init() {
    read(n), read(m), read(k);
    for (int i = 1; i <= m; ++i) {
        int x, y, z;
        read(x), read(y), read(z);
        if (z == 1) ++A.a[x][y];
        if (z == 2) ++A.a[x][y + n];
        if (z == 3) ++A.a[x][y + n * 2];
    }
    for (int i = 1; i <= n; ++i) A.a[i][0] = A.a[i + n][i] = A.a[i + n * 2][i + n] = 1;
    A.a[0][0] = 1;
}

int main() {
#ifdef hzhkk 
    freopen("hkk.in", "r", stdin);
#endif
    init();
    work();
    fclose(stdin), fclose(stdout);
    return 0;
}

原文地址：https://www.cnblogs.com/hankeke/p/BZOJ4386.html